Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{4r - 4}{8r + 40} \times \dfrac{r^2 - 25}{3r - 3} $
Answer: First factor the quadratic. $k = \dfrac{4r - 4}{8r + 40} \times \dfrac{(r + 5)(r - 5)}{3r - 3} $ Then factor out any other terms. $k = \dfrac{4(r - 1)}{8(r + 5)} \times \dfrac{(r + 5)(r - 5)}{3(r - 1)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ 4(r - 1) \times (r + 5)(r - 5) } { 8(r + 5) \times 3(r - 1) } $ $k = \dfrac{ 4(r - 1)(r + 5)(r - 5)}{ 24(r + 5)(r - 1)} $ Notice that $(r - 1)$ and $(r + 5)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ 4\cancel{(r - 1)}(r + 5)(r - 5)}{ 24\cancel{(r + 5)}(r - 1)} $ We are dividing by $r + 5$ , so $r + 5 \neq 0$ Therefore, $r \neq -5$ $k = \dfrac{ 4\cancel{(r - 1)}\cancel{(r + 5)}(r - 5)}{ 24\cancel{(r + 5)}\cancel{(r - 1)}} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $k = \dfrac{4(r - 5)}{24} $ $k = \dfrac{r - 5}{6} ; \space r \neq -5 ; \space r \neq 1 $